# What is the answer for question 8 using question number 7. b) http://postimg.org/image/sx71ajgbb/

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You need to evaluate the inverse of the function `f(x) = 1/(x - 2) + 3` , such that:

`y = 1/(x - 2) + 3 => y - 3 = 1/(x - 2) => (x - 2)(y - 3) = 1 => x - 2 = 1/(y - 3) => x = 1/(y - 3) + 2`

Hence, evaluating the function inverse yields` f^(-1)(x) = 1/(x - 3) + 2`

You need to sketch the graphs of the functions `f(x) = 1/(x - 2) + 3` and `f^(-1)(x) = 1/(x - 3) + 2` , such that:

The black curve represents the function ` y = 1/(x - 2) + 3 ` and the red curve represents the graph of its inverse.

**You may notice that the domain of the function `y = 1/(x - 2) + 3` is R, except `x = 2` and the domain of the function inverse `y = 1/(x - 3) + 2` is R, except `x = 3` .**

You need to evaluate the function inverse of `y = 2(x - 4)^2 + 5` , such that:

`y = 2(x - 4)^2 + 5 => y - 5 = 2(x - 4)^2 => (x - 4)^2 = (y - 5)/2 => x - 4 = +-sqrt((y - 5)/2) => x = 4 +- sqrt((y - 5)/2)`

Hence, evaluating the function inverse yields `f^(-1)(x) = 4 +- sqrt((x - 5)/2)`

You should notice that the domain of the function contains the values of x that make the square root to hold, such that:

`(x - 5)/2 >= 0 => x - 5 >= 0 => x >= 5`

Hence, the domain of the inverse function is `[5,oo)` , while the domain of function is R and the range is `[5,oo)` . You should notice that the domain of the function represents the range of its inverse and the range of the function represents the domain of its inverse (you may obtain the graph of the inverse rotating clockwise to `90^o` the black curve, that represents the graph of function).